1. Set Up the Angular Accelerations ($\alpha = \frac{\tau}{I}$)

Let $l$ be the length of the light rigid rod. We calculate the initial angular acceleration $\alpha$ at any arbitrary angle $\theta$ with the vertical using $\tau = I\alpha$.

Case 1: Single bead of mass $m$ at the tip ($l$)

$$ \alpha_1 = \frac{\tau_1}{I_1} = \frac{mgl \sin\theta}{ml^2} = \left(\frac{g}{l}\right)\sin\theta $$

Case 2: Additional bead of mass $\eta m$ at the midpoint ($l/2$)

$$ \alpha_2 = \frac{\tau_2}{I_2} = \frac{mgl \sin\theta + \eta mg \left(\frac{l}{2}\right) \sin\theta}{ml^2 + \eta m \left(\frac{l}{2}\right)^2} $$

Factoring out the common terms:

$$ \alpha_2 = \frac{mgl\left(1 + \frac{\eta}{2}\right)\sin\theta}{ml^2\left(1 + \frac{\eta}{4}\right)} = \left(\frac{g}{l}\right) \left( \frac{1 + \frac{\eta}{2}}{1 + \frac{\eta}{4}} \right) \sin\theta $$

2. The Scaling Property

Notice that both equations share the exact same functional form: $\alpha = K \sin\theta$. Because their paths are geometrically identical, their characteristic frequencies scale as $\omega \propto \sqrt{K}$, meaning the total time taken relates inversely to the square root of the scaling constant:

$$ t \propto \frac{1}{\sqrt{K}} $$

3. Direct Ratio Calculation

We can now directly compare the new time $t_{new}$ to the original time $t_0$ by comparing their acceleration coefficients:

$$ \frac{t_{new}}{t_0} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{1}{\frac{1 + \eta/2}{1 + \eta/4}}} = \sqrt{\frac{1 + \frac{\eta}{4}}{1 + \frac{\eta}{2}}} $$

Simplifying the algebraic fraction:

$$ \frac{t_{new}}{t_0} = \sqrt{\frac{\frac{4+\eta}{4}}{\frac{2+\eta}{2}}} = \sqrt{\frac{4+\eta}{2(2+\eta)}} $$

Thus, the final time expression is:

$$ t_{new} = t_0 \sqrt{\frac{\eta + 4}{2(\eta + 2)}} $$

1. Derivation of Time Dependence ($t \propto 1/\sqrt{K}$)

Consider a rigid body pivoted at the bottom, falling under gravity. The equation of motion is given by Torque = $I \alpha$.
Let $\phi$ be the angle with the vertical at any instant (varying from initial $\theta$ to $90^\circ$). The restoring torque due to gravity is proportional to $\sin\phi$. $$ \tau = I \alpha \implies \tau_{eff} \sin\phi = I \frac{d^2\phi}{dt^2} $$ $$ \frac{d^2\phi}{dt^2} = \left( \frac{\tau_{eff}}{I} \right) \sin\phi $$ Let $K = \frac{\tau_{eff}}{I}$. Then: $$ \ddot{\phi} = K \sin\phi $$ To integrate, we multiply both sides by $2\dot{\phi}$: $$ 2\dot{\phi} \frac{d\dot{\phi}}{dt} = 2K \sin\phi \frac{d\phi}{dt} $$ $$ \int d(\dot{\phi}^2) = 2K \int \sin\phi \, d\phi $$ Assuming the rod starts from rest at angle $\theta$: $$ \dot{\phi}^2 = 2K (\cos\theta – \cos\phi) $$ $$ \frac{d\phi}{dt} = \sqrt{2K} \sqrt{\cos\theta – \cos\phi} $$ Separating variables to find time $t$: $$ dt = \frac{1}{\sqrt{2K}} \frac{d\phi}{\sqrt{\cos\theta – \cos\phi}} $$ Integrating from release angle $\theta$ to the ground ($\phi = \pi/2$): $$ t = \frac{1}{\sqrt{2K}} \int_{\theta}^{\pi/2} \frac{d\phi}{\sqrt{\cos\theta – \cos\phi}} $$ The integral term depends only on the geometry (angles) and is constant for both cases. Therefore: $$ t \propto \frac{1}{\sqrt{K}} $$

2. Calculating Factor K for Both Cases

Case 1: Single Mass $m$ at length $l$
Torque about pivot: $\tau_1 = mgl$. Moment of Inertia: $I_1 = ml^2$. $$ K_1 = \frac{mgl}{ml^2} = \frac{g}{l} $$

Case 2: Added Mass $\eta m$ at length $l/2$
Total Torque (coefficient of $\sin\phi$): $$ \tau_2 = mgl + (\eta m)g\left(\frac{l}{2}\right) = mgl \left(1 + \frac{\eta}{2}\right) $$ Total Moment of Inertia: $$ I_2 = ml^2 + (\eta m)\left(\frac{l}{2}\right)^2 = ml^2 \left(1 + \frac{\eta}{4}\right) $$ Calculating $K_2$: $$ K_2 = \frac{\tau_2}{I_2} = \frac{mgl(1 + \eta/2)}{ml^2(1 + \eta/4)} = \frac{g}{l} \left( \frac{1 + \eta/2}{1 + \eta/4} \right) $$

3. Comparison of Times

Using the relation $t \propto 1/\sqrt{K}$: $$ \frac{t_{new}}{t_0} = \sqrt{\frac{K_1}{K_2}} $$ Substitute $K_1$ and $K_2$: $$ \frac{t_{new}}{t_0} = \sqrt{ \frac{g/l}{\frac{g}{l} \left( \frac{1 + \eta/2}{1 + \eta/4} \right)} } = \sqrt{ \frac{1 + \eta/4}{1 + \eta/2} } $$ Simplifying the fraction: $$ \frac{1 + \eta/4}{1 + \eta/2} = \frac{(4+\eta)/4}{(2+\eta)/2} = \frac{4+\eta}{2(2+\eta)} $$ $$ t_{new} = t_0 \sqrt{\frac{\eta+4}{2(\eta+2)}} $$